A 25.02-ml sample of piperazine (a diprotic acid, H2Pip) was titrated with 0.010

Question

A 25.02-ml sample of piperazine (a diprotic acid, H2Pip) was titrated with 0.01075 M NaOH. A 50.00-ml portion of the NaOH was added, which was enough to overshoot the second equilibrium point. Any remaining NaOH in solution was then titrated requiring 7.05-ml of standardized solution of HNO3 with a concentration of 0.02478 M. [Ka1=4.65x10^-6 ; Ka2= 1.86x10^-10]

A. What is the concentration of piperazine in the original sample?

B. Given the values for Ka above, estimate the pH at which the concentration of the amphiprotic form of piperazine would be highest in solution

C. Write out the second base dissociation reaction and determine the value for Kb2 for this acid/conjugate base system.

Explanation / Answer

Let’s consider that,

Initially for H2Pip, Molarity = M1 = ? and volume = 25.02 mL

For NaOH, M2 = 0.01075 M and V2 = 50.00 mL

Initial milimoles of NaOH = M2 x V2 = 0.5375 milimoles NaOH.

For HNO3 M3 = 0.02478 M and V3 = 7.05 mL

As per given information,

Milimoles of HNO3 = M3 x V3 = 0.02478 x 7.05 = 0.1747 milimoles.

Hence its clear that thses are the excess milimoles of NaOH left on treatment of NaOH with H2Pip.

Excess milimoles of NaOH = 0.1747

The actual milimoles of NaOH required to neutralize H2Pip completey is given as,

Initial milimoles of NaOH – excess milimoles of NaOH

= 0.5375 – 0.1747

= 0.3628 milimoles.

I.e. Actual milimoles of NaOH that required for complete neutralization of H2Pip is 0.3628 milimoles.

But this milimoles are the milimoles of H2Pip solution.

Milimoles of H2pip = 0.3628

½ M1 x V1 = 0.3628

M1 x V1 = 2 x 0.3628

M1 x 25.02 = 0.7256

M1 = 0.029 M

In original sample concentration of H2Pip was 0.029 M.

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B)

H2Pip ionization is shown as,

First ionization:   H2Pip ------------> HPip– + H+ (aq.)   Ka1 = 4.56 x 10–6

Second Ionization: HPip– -------- > Pip2– + H+ (aq.)    Ka2 = 1.86 x 10–10

But Q.B asks to find H+ concentration when amphiprotic species HPip– is maxium and its obvious that its after first ionization step only hence,

Initially for H2Pip we have M1 = 0.029 M and V1 = 25.02 mL = 25.02 x 10–3 L

Hence, [H2Pip] = 0.029 x 25.02 x 10–3 = 7.26 x 10–4.

Corresponding Ionization and ICE table is

First ionization:   H2Pip ------------>      HPip– + H+ (aq.)   Ka1 = 4.56 x 10–6

Initially              7.26 x 10–4.                0            0

Change                    –X                         X              X

At eqm.              (7.26 x 10–4 – X)      ‘X’ M          ‘X’ M

Expression for equilibrium constant,

Ka1 = [HPip–][H+]/[H2Pip]

By ICE table

4.56 x 10–6 = (X) (X) /(7.26 x 10–4 – X)

By small concentration assumption we write,

4.56 x 10–6 = (X) (X) /(7.26 x 10–4 –

X2 = 4.56 x 10–6 x 7.26 x 10 –4

X2 = 3.31 x 10–9.

X = 5.75 x 10–5

By ICE table,

[H] = 5.75 x 10–5 M/L

Hence, pH = –log(5.75 x 10–5)

pH = 4.24

pH corresponds to the condition at which amphiprotic species is highest in concentration is 4.24.

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C) Second base dissociation reaction is,

HPip– -------- > Pip2– + H+ (aq.)    Ka2 = 1.86 x 10–10

We know that,

Ka x Kb = Kw = 1 x 10–14

Hence for above ionization we can write,

Kb2 x Ka2 = 1 x 10–14

We know Ka2 as Ka2 = 1.86 x 10–10 using this value we have,

Kb2 x 1.86 x 10–10   = 1 x 10–14

Kb2 = 1 x 10–14/ 1.86 x 10–10 .

Kb2 = 5.38 x 10–5.

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